Question

A salt container is to be made in the shape of a cylinder and to contain `500cm^3` of salt. Find the dimensions for the container that requires the least amount of material. Assume the container has a top?

Answer 1

The smallest area occurs when we have a radius of `4.30 cm` (2dp), leading to an area of `A=348.73 cm^3` (2dp).

Explanation:

Let us set up the following variables:

`{(r, "Radius (cm"), (h, "Height of can (cm)"), (A, "Surface Area of can (sq cm)") :}`

We want to vary the radius `r` and height `h` such that we minimise `A`, ie find a critical point of `A` wrt `r` (wlog), ie find `(dA)/(dr)` that is a minimum, so we to find a function `A(r)`

Then the volume is fixed:

`pir^2h = 500`
`:. h = 500/(pir^2)`

And, The Surface Area is given by:

`\ \ \ \ \ A=2pirh + 2pir^2`
`:. A=2pir(500/(pir^2)) + 2pir^2`
`:. A=1000/r + 2pir^2`

Differentiating wrt `r` gives us;

`:. (dA)/(dr)=(-1000/r^2) + 4pir`

At a critical point, `(dA)/(dr)=0`

`:. (-1000/r^2) + 4pir = 0`
`:. -1000 + 4pir^3 = 0`
`:. r^3 = 1000/(4pi)`
`:. r = root(3)(1000/(4pi))`
`:. r = 4.30 (2dp)`

With `r=4.30` we have:

`A=1000/(4.30) + 2pi(4.30)^2`
`:. A=348.73 (2dp)`

We should check that this value leads to a minimum (rather than a maximum) area. As the size of the can is finite this should really be intuitive. We could calculate the second derivative and verify that `(d^2A)/(dr)^2 > 0` when `r=1.2` Instead I will just use the graph `A=(6.912pi)/r + 2pir^2`

graph{1000/x + 2pix^2 [-15, 10, 200, 490.7]}

Hopefully you can visually confirm that a minimum does indeed occur when `4.30 (2dp)`